Interpolation percolation

Let X be a countably infinite set of real numbers and let Y_x, x \in X, be an independent family of stationary random subsets of the real numbers, e.g. homogeneous Poisson point processes. We give criteria for the a.s. existence of various"regular"functions f with the property that f(x) \in Y_x for all x \in X. Several open questions are posed.


Introduction
In classical discrete percolation [Gr99] one randomly and independently deletes edges or vertices from a graph and considers the properties of the connected components of the remaining graph V . In standard continuum percolation [MR96], which is concerned with Boolean models, one removes balls, whose centers form a random point process, from space and again investigates the connectivity properties of the remaining part V of space or, somewhat more commonly, of its complement V c . In both cases, each component deleted from the underlying medium has, in some sense, a strictly positive volume. Moreover, any bounded region of the space intersects a.s. only a finite number of these components (see e.g. [MR96,Proposition 7.4] for continuum percolation).
In contrast, in fractal percolation [Gr99,Chapter 13.4] and continuum fractal percolation [Zä84] [MR96,Chapter 8.1] one may delete from any bounded open region countably many components, each of which has a strictly positive volume. For example, [Zä84] deals with the Hausdorff dimension of the set V = R n \ i∈N Γ i , where Γ 1 , Γ 2 , . . . are independent random open sets in R n , the so-called cutouts, e.g. scaled Boolean models.
In the present paper we introduce another continuum percolation model, in R 2 , in which countably many components may be removed from any bounded region. However, in contrast to the previous models, these components are null sets in R 2 . In particular, the remaining set V ⊂ R 2 has full Lebesgue measure. The cutouts are arranged in such a way that V exhibits various phase transitions. The precise model is the following.
Fix a complete probability space (Ω, F , P ) and a countably infinite set X ⊂ R. For all x ∈ X let Y x ⊆ R be a random closed set, i.e. Y x is a random variable on (Ω, F , P ) with values in the space of closed subsets of R equipped with the Borel σ-algebra generated by the so-called Fell topology, see [Mo05, Chapter 1.1.1] for details. Throughout we suppose that Another common assumption will be that which is defined as usual, see [Mo05, Definition 1.1.18]. We will also need to assume some translation invariance of the sets Y x as described in [Mo05, Chapter 1.4.1]. One such possible assumption is that (STAT) Y x is stationary for all x ∈ X , i.e. Y x has for all a ∈ R the same distribution as Y x + a.
This assumption is stronger than the hypothesis that (1STAT) Y x is first-order stationary for all x ∈ X , i.e. for all a, b, c ∈ R, Our main example, which satisfies all of the above conditions, is the following: (PPP) (Y x ) x∈X are independent homogeneous Poisson point processes on R with intensities λ x > 0, where (λ x ) x∈X is fixed.
For a renewal process with interarrival times which are not exponentially distributed, like under (PPP), but Weibull distributed, see Example 4. Another example, which fulfills (IND) and (STAT) and is periodic, is (2) Y x = λ x (K x + Z + U x ), where λ x = 0 and all K x ⊂ R (x ∈ X ) are compact and (U x ) x∈X is i.i.d. with U x ∼ Unif[0, 1].
We now remove for each x ∈ X the set {x} × Y c x from R 2 , where Y c x = R\Y x , and investigate the remaining set We thus "perforate" R 2 by cutting out random subsets of parallel vertical lines to obtain a "vertically dependent" random set V ⊂ R 2 . For a similar discrete percolation model with vertical dependence see [Gr09, Section 1.6].
What are the topological properties of V ? It is easy to see that V is always connected, see Proposition 5. However, whether V is path-connected or not depends on the parameters.  Note that (3) depends only on the intensities λ x , counted with multiplicities, but not on X itself.
Theorem 1 will be generalized in Theorem 4. There it will also be shown that V is a.s. path-connected if and only if there is a continuous function f : R → R whose graph graph(f ) : This brings up the question under which conditions there are functions f : R → R which have other regularity properties than continuity and which belong to The elements of I in some sense interpolate the sets Y x , x ∈ X , see Figure 1 for examples. For this reason we suggest the name interpolation percolation for this model. In Section 2 we derive some conditions under which the following subsets of Ω, which describe the existence of interpolating functions with various regularity properties, occur or don't occur. It will follow from the completeness of (Ω, F , P ) that these sets are events, i.e. elements of F . Our results for B, P m and A are not difficult.
Remark 1. (0-1-law) Note that every event defined above, call it G, is invariant under vertical shifts of V , i.e. G occurs if and only if it occurs after replacing V by any V + (0, a), a ∈ R. Therefore, under suitable ergodicity assumptions, . This is the case if (PPP) holds. In general, e.g. in case (2), this need not be true, see Remark 10.

Remark 2. (Monotonicity)
There is an obvious monotonicity property: If X is replaced by for all x ′ ∈ X ′ then V and I and all the events defined above and their probabilities increase.
Remark 3. (Closedness) There are various notions of random sets. Random closed sets seem to be the best studied ones, see [Mo05] and Chapter 1.2.5 therein for a discussion of non-closed random sets. For this reason we assume the sets Y x to be closed even though this assumption does not seem to be essential for our results. An alternative, but seemingly less common notion of stationary random and not necessarily closed sets is described e.g. in [JKO94,Chapter 8]. (a) (Brownian motion) Our construction of the continuous functions in the proof of Theorem 1 has been inspired by Paul Lévy's method of constructing Brownian motion. In Example 4 we shall even choose X ⊆ (0, 1] and (Y x ) x∈X in a nontrivial way and use this method to define a Brownian motion (B x ) 0≤x≤1 on the same probability space (Ω, F , P ) such that a.s. (B x ) x∈R ∈ I. (Here we let B x := 0 for x / ∈ (0, 1] to extend B · (ω) to a function defined on R.) (b) (Lipschitz and directed percolation) In [DDGHS10] random Lipschitz functions F : Z d−1 → Z are constructed such that, for every x ∈ Z d−1 , the site (x, F (x)) is open in a site percolation process on Z d . The case d = 2 is easy since it is closely related to oriented site percolation on Z 2 . However, if we let d = 2, denote the Lipschitz constant by L and let the parameter p L of the site percolation process depend on L such that Lp L → λ ∈ (0, ∞) as L → ∞ then we obtain after applying the scaling (x, y) → (x, y/L) in the limit L → ∞ the problem of studying the event L 1 under the assumption (PPP) with X = Z and λ x = λ for all x ∈ X . Theorem 10 deals with this case and is proved using oriented percolation.
(c) (First-passage percolation) Our computation of P [BV ] in Theorem 7 applies a method used for the study of first-passage percolation on spherically symmetric trees in [PP94].
(d) (Intersections of stationary random sets) Note that Such intersections of stationary random sets (or the unions of their complements) have been investigated e.g. in [Sh72a], [Sh72b], [KP91] and [JS08]. Studying the events P m ⊇ P 0 , m ≥ 1, yields natural variations of these problems, see e.g. Open Problem 3. (e) (Poisson matching) In 2-color Poisson matching [HPPS09] one is concerned with matching the points of one Poisson point process to the points of another such process in a translation-invariant way. In our model with assumption (PPP), ∞-color Poisson matching would correspond to choosing infinitely many f ∈ I in a translation-invariant way. This will be made more precise in Open Problem 6.

Results, proofs, and open problems
Some of our results will be phrased in terms of the following random variables. We denote for x ∈ X and z ∈ R by Lemma 2. The probability that D x (z) is finite for all x ∈ X and all z ∈ R is 1. Similarly, Example 2. Under assumption (PPP) all three distances D + x (z), D − x (z) and D x (z) are exponentially distributed with respective parameters λ x , λ x , and 2λ x .
For comparison with Theorem 4 about continuous functions and as a warm-up we first consider bounded, but not necessarily continuous functions.
Proof. If (5) holds then a bounded function f ∈ I can be defined by setting For the converse we note that any f ∈ I must satisfy f ∞ ≥ sup x∈X D x (0) and apply Kolmogorov's zero-one law.
Remark 5. By the Borel Cantelli lemma (5) is equivalent to Note that (5) and (6) do not depend on X but only on the distributions of the random variables D x (0), x ∈ X . Theorem 1 immediately follows from the following result.

Theorem 4. (Continuous functions) Assume (IND) and (1STAT). If
Remark 6. Suppose (x n ) n∈N enumerates X . Then, by the Borel Cantelli lemma, (7) is equivalent to Compare (7) to (6) and (8) to (5). Also note that if X is bounded then (7) does not depend on X itself but only on the distributions of the random variables D x (0), x ∈ X .
Proof of Theorem 4. Assume (7). Then for all y ∈ R and all ε > 0 the set X (y, ε) := {x ∈ X | D x (y) ≥ ε/3} is a.s. locally finite due to the Borel Cantelli lemma and Lemma 2. Therefore, for all K, ε > 0 the set is a.s. locally finite as well since it is contained in the union of the sets X (y, ε) with y ∈ [−K, K] ∩ (ε/3)Z. This together with (1) and the monotonicity of X K (ε) in K and in ε implies the existence of a set Ω ′ ⊆ Ω of full P -measure on which For the proof of the first statement of the theorem it suffices to show that on Indeed, this immediately shows that on Ω ′ any two points in V with differing first coordinates a 0 and a 1 can be connected by a continuous path inside V . Points in V whose first coordinates coincide can be connected by the concatenation of two such paths. Similarly, P [C] = 1 follows by concatenating the functions f an,a n+1 ,0,0 , where (a n ) n∈Z is a strictly increasing double sided sequence in R\X with a n → ∞ and a −n → −∞ as n → ∞. Therefore, let a 0 , a 1 , b 0 , b 1 ∈ R with a 0 < a 1 . If X ∩]a 0 , a 1 [ is finite then the existence of a continuous function f a 0 ,a 1 ,b 0 ,b 1 satisfying (10) is obvious. Now assume that X ∩]a 0 , a 1 [ is infinite. Also fix a realization in Ω ′ . By (9), is finite. Here sup ∅ := 0. The sets A n , n ∈ N 0 , recursively defined by are finite due to (9), pairwise disjoint and satisfy We shall obtain the desired function f = f a 0 ,a 1 ,b 0 ,b 1 as uniform limit of a sequence (f n ) n≥0 of continuous functions which satisfy for all n ≥ 0, We construct this sequence recursively. For . This defines f 0 (x) for finitely many x. By linearly interpolating in between these values we get a continuous piecewise linear function f 0 satisfying (13)-(17) with n = 0. Now let n ≥ 0 and assume that we have already constructed a continuous function f n which fulfils (13)-(17). We then set For all x ∈ A n+1 , which again is a finite set, we choose f n+1 (x) as some element of Y x with distance D x (f n (x)) from f n (x) and again interpolate linearly to obtain a piecewise linear continuous function f n+1 , see Figure  2 (a). Obviously, this function satisfies (14)-(16) with n + 1 instead of n. To see that the same is true for (13) and (17) we note that for all x ∈ A n+1 , by i.e. (17) holds with n + 1 instead of n. This together with (13) implies that (13) also holds for n + 1 instead of n.
Having finished the construction of (f n ) n≥0 , we see that it converges uniformly on [a 0 , a 1 ] due to (17) to some function f . Since all the functions f n are continuous f is continuous as well. It also fulfils (10) due to (12)-(16).
For the proof of the converse assume that there are m ∈ N and ε > 0 such that Without loss of generality we may assume that one can choose ξ ∈ [−m, m] such that even (Otherwise replace X by −X .) Denote by C 2 the event that there is a con- Figure 2 (b). Therefore, f 1 (σ f ) < ξ by definition of τ f . Consequently, For such t we have on the one hand by the definition of σ f that |f 2 (t) − f 2 (τ f )| < ε/2 and on the other hand f 2 (t) ∈ Y x since graph(f ) ⊂ V . Therefore, by the triangle inequality, Consequently, it suffices to show that P [C j,k ] = 0 for all j ∈ Z, k ∈ N. However, by (IND), (1STAT) and Lemma 2, Whether V is path-connected or not thus depends on the parameters of the model. One may wonder whether the same is true for the connectedness of V . This is not the case. V is always connected as the following non-probabilistic statement shows when applied to U = V and X = R\X .   We shall show that one can indeed choose X ⊆ (0, 1] and, in a non-trivial way, (Y x ) x∈X and then construct by Lévy's method a Brownian motion (B x ) x∈[0,1] on (Ω, F , P ) such that a.s. (B x ) x∈R ∈ I. (Here B x := 0 for x / ∈ [0, 1].) Like in Lévy's construction we let X be the set of dyadic numbers in (0, 1], namely the disjoint union X := X −1 ∪ n≥1 X n , where X −1 := {1} and X n := k2 −n 1 ≤ k < 2 n , k is odd for n ≥ 1. For n = −1, 1, 2, 3, . . . and x ∈ X n let Y x ⊂ R be a twosided stationary (in the sense of (STAT), see e.g. [KT75, Theorem 9.9.1]) renewal processes with i.i.d. interarrival times whose cumulative distribution function is given by F n (t) := 1 − exp (−2 n−2 t 2 ) for t ≥ 0 and whose probability density function we denote by f n . Furthermore, we assume (IND). For x ∈ X and z ∈ R we denote by Y x (z) the a.s. unique element of Y x with minimal distance to z, i.e. distance D x (z). We then recursively define B x for x ∈ {0} ∪ X as follows: We set B 0 := 0 and B 1 := Y 1 (0). Having defined B x for x ∈ X m , m < n, we set for x ∈ X n . By Lemma 2 and [KT75, Theorem 9.9.1] for all x ∈ X n and z ∈ R, To prove the first assertion assume that (D + x (0)) x∈X is a.s. summable. For y ∈ R and n ≥ 0 we denote by u n (y) := y + D + xn (y) the smallest element of Y xn which is Firstly, f is increasing since each f n is increasing. Secondly, f is a.s. bounded. Indeed, due to (IND), (1STAT) and Lemma 2, f n ∞ = f n (x ϕn(n) ) has the same distribution as D + x 0 (0) + . . . + D + xn (0). Moreover, (f n (x)) n≥0 is for each x ∈ R an increasing sequence. Consequently, f n ∞ increases for n → ∞ towards f ∞ . Hence f ∞ has the same distribution as x∈X D + x (0) and is therefore a.s. finite by assumption. Thirdly, we have for all For the proof of the second statement of the theorem assume that Note that M is the union of all M a,b , a < b. Therefore, it suffices to show that P [M a,b ] = 0 for all a < b. For all n ≥ 0, which is, by assumption, strictly less than 1 and therefore, by Kolmogorov's zeroone law, equal to 0.
Open Problem 1. (a) Theorem 7 and Example 5 do not determine P [BV ] in some cases in which (λ n ) n≥0 is a permutation of, say, (n log n) n≥2 . (b) Can Theorem 7 be extended to X whose closure contains an interval?
The following non-probabilistic lemma is needed for case (24). For the proof of Theorem 7 we first show in Lemma 9 that whenever there is a function of bounded variation in I then there is also another such function which is only "jumping between nearest neighbors". More precisely, we recursively define the random function g : n∈N 0 {+, −} n → R such that g(s 1 , . . . , s n ) ∈ Y n for n ≥ 1 by setting g and set s := (s i ) i∈N . Thus (g i ) i≥0 := g s is the sequence which starts at 0 and tries to trace (f i ) i≥0 but is restricted to making only the smallest possible jumps up or down. To prove V (g s ) < ∞ we consider for all n ∈ N the telescopic sum We shall show that all the summands in (32) are non-negative, see Figure 4. Since the sequences (g 0 , . . . , g i−1 , f i , . . . , f n ) and (g 0 , . . . , g i , f i+1 , . . . , f n ) differ only in the i-th term the i-th summand in (32) is equal to where we used in (33) that by definition Due to the triangle inequality the expression in (33) is non-negative. Hence, by (32), V (g 0 , . . . , g n ) ≤ V (f 0 , . . . , f n ) and therefore which is less than or equal to the total variation of f , which is finite. This is the sum of independent exponentially distributed random variables with respective parameter λ i , see Example 2. To apply standard large deviation estimates we weight the summands to make them i.i.d. and consider first the sum (s 1 , . . . , s i−1 )) of n independent random variables which are all exponentially distributed with parameter 1. We choose ε > 0 such that 2e 2ε < 3, denote by + n the only element of {+} n and get P min s∈{+,−} n V * (g s ) < εn ≤ 2 n P [V * (+ n ) < εn] ≤ 2 n e 2εn E e −2V * (+ n ) = 2e 2ε 3 n , which is summable in n. Therefore, by the Borel Cantelli lemma, there is a.s. some random N such that (37) M * n := min s∈{+,−} n V * (g s ) ≥ εn for all n ≥ N.
Recalling (35) we have for all n ∈ N and s ∈ {+, −} n , (s 1 , . . . , s j−1 )) Due to assumption (22), λ −1 i − λ −1 i+1 ≥ 0. Therefore, the right hand side in (38) is The second summand in (39) is equal to Therefore, the right hand side of (39) is equal to The right hand side of (40) does not depend on the particular choice of s ∈ {+, −} n and tends a.s. to ∞ as n → ∞ due to (22). Consequently, the right most side of (34) is 0, which completes the proof in case (22). In case (23) there is some finite constant c such that X ′ := {x ∈ X : λ x < c} is infinite. Since X ′ ⊆ X and (λ x ) x∈X ′ ≤ (c) x∈X ′ the claim follows from (22) and monotonicity (Remark 2).
Case (24) is treated similarly. We choose a permutation ϕ of N such that λ (n) = λ ϕ −1 (n) for all n. Applying Lemma 8 to µ n := 1/λ (n) gives a set X ′ : The statement now follows from case (22) by monotonicity (Remark 2).  Open Problem 2. What is the exact value of λ c ? Critical thresholds in (oriented) percolation are rarely explicitly known. However, due to Monte-Carlo simulations, see also Figure 5 (b), we conjecture λ c = 1.
In the following simple example the corresponding λ c can be computed explicitly.
Example 6. Let X = Z and λ, K > 0. Let (U n ) n∈Z be independent and uniformly distributed on [0, 1] and set Y n := (U n + Z n )/λ for n ∈ Z. (Note the similarity of V to perforated toilet paper.) Then for any n ∈ Z and z ∈ R there is some y ∈ Y n with |z − y| ≤ K if 2K ≥ 1/λ. If 2K < 1/λ then there is at most one such y and with positive probability no such y. Hence P [L K ] = 1 if λ ≥ 1/(2K) and P [L K ] = 0 else. Thus in this case (42) holds for λ c = 1/2.
Next we consider some of the smoothest functions, polynomials and real analytic functions. Proof. Let m ∈ N 0 and let x 0 , . . . , x m+1 ∈ X be pairwise distinct. For any y = (y 0 , . . . , is countable and compact. Hence, } is a.s. countable and, by continuity, compact as well. If P m occurs then G K ∩ Y x m+1 = ∅ for some K ∈ N. By (IND) and (STAT), the closed set G K ∩ Y x m+1 has for all a ∈ R the same distribution as G K ∩ (Y x m+1 + a). Therefore, Using Fubini's theorem and denoting by λ the Lebesgue measure on R, we get Of course, the assumption of countability in Proposition 11 cannot be dropped. Here is a nontrivial example. Let (x n ) n∈N enumerate X and let 1 > ℓ 1 ≥ ℓ 2 ≥ . . . ≥ ℓ n → 0 as n → ∞. We consider two different families (Y x ) x∈X . The first one is of the type described in (2) and is given by cf. Figure 1 (b). The second one consists of complements of Boolean models: are independent Poisson point processes with intensity 1.
In case (43), P [0 ∈ I] = n≥1 (1 − ℓ n ), while in case (44), P [0 ∈ I] = n≥1 e −ℓn . Therefore, in either case P [0 ∈ I] = 0 if and only if n ℓ n = ∞. Obviously, {0 ∈ I} ⊆ P 0 . However, the following theorem due to L. A. Shepp shows that these two events might differ by more than a null set. To recognize it recall (4) and take complements. Remark 10. Note that in case (43) there is no zero-one law like in Remark 1: 0 < P [P 0 ] < 1 is possible. In fact, P [P 0 ] = 1 if and only if n ℓ n ≤ 1, which is not the opposite of (45).
For m = 1 and case (43) (see also Figure 1 (b)) this problem can be phrased in terms of random coverings of a circle in the spirit of [Sh72a] and [JS08] as follows: Arcs of length ℓ x (x ∈ X ) are thrown independently and uniformly on a circle of unit length and then rotate at respective speed x around the circle. Give a necessary and sufficient condition in terms of (ℓ x ) x∈X and X under which there is a.s. no (resp., a.s. at least one) point in time at which the circle is not completely covered by the arcs. In other words: Under which conditions is there a.s. no (resp., a.s. at least one) random, but constant speed at which one can drive along a road with infinitely many independent traffic lights without ever running into a red light?
Proposition 13. (Real analytic functions) If X is locally finite then P [A] = 1.
Proof. Choose y x ∈ Y x for all x ∈ X . By [Ru87,Theorem 15.13] there is an entire function g : C → C, g(z) = n≥0 a n z n , such that g(x) = y x for all x ∈ X . Its real part ℜ(g(z)) = n≥0 ℜ(a n )z n restricted to R is real analytic and takes values y x at x ∈ X as well.
We conclude by suggesting some further directions of research.
Open Problem 4. Theorem 10 and Proposition 13 deal only with locally finite X . What can be said about P [L K ] and P [A] for more general X ? It is easy to see that even in case (PPP) with constant intensities λ x = λ any general criterion for, say, P [L K ] = 0 would need to depend not only on λ but also on X itself. If X is for example bounded then for any K > 0, P [L K ] ≤ P [C] = 0 by Theorem 4, no matter how large λ is, in contrast to (42).
Open Problem 5. One might consider other types of interpolating functions. For example, under which conditions are there functions f ∈ I, which are (a) continuous and monotone at the same time or (b) Hölder continuous or (c) k-times continuously differentiable? Extensions to higher dimension might be possible as well.
Open Problem 6. (More than just one function) Let X ⊂ [0, 1] and fix (λ x ) x∈X with λ x > 0. Under which conditions is there a simple point process N = i∈N δ f i (in the sense of [DV88, Definition 7.1.VII]) on the space C([0, 1]) of continuous functions on [0, 1] (or any other suitable space of regular functions) such that (a) for all x ∈ X the points f i (x), i ∈ N, are pairwise distinct, (b) there are independent homogeneous Poisson point processes Y x , x ∈ X , with intensities λ x such that {f i (x) : i ∈ N} ⊆ Y x for all x ∈ X and (c) the "vertically shifted" point process i∈N δ f i +y has for all y ∈ R the same distribution as N?